Questions: Solve by the elimination method
5x - y = 13
x + 6y = 15
Select the correct choice below and, if necessary, fill in the answer box to complete your choice
A. The solution is . (Simplify your answer. Type an ordered pair.)
B. There are infinitely many solutions.
C. There is no solution.
Transcript text: Solve by the elimination method
\[
\begin{array}{c}
5 x-y=13 \\
x+6 y=15
\end{array}
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choi
A. The solution is $\square$ I.
(Simplify your answer. Type an ordered pair.)
B. There are infinitely many solutions.
C. There is no solution.
Solution
Solution Steps
To solve the system of equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the equations. We can multiply the equations by suitable numbers to make the coefficients of one of the variables equal in magnitude but opposite in sign. Then, we add or subtract the equations to eliminate that variable and solve for the remaining variable. Finally, we substitute back to find the other variable.
Step 1: Set Up the Equations
We start with the system of equations:
\[
\begin{align*}
& \quad 5x - y = 13 \\
& \quad x + 6y = 15
\end{align*}
\]
Step 2: Solve for One Variable
Using the elimination method, we can manipulate the equations to eliminate \(y\). We can multiply the first equation by \(6\) to align the coefficients of \(y\):
\[
6(5x - y) = 6(13) \implies 30x - 6y = 78
\]
Now we have:
\[
\begin{align_}
3) & \quad 30x - 6y = 78 \\
2) & \quad x + 6y = 15
\end{align_}
\]
Step 3: Add the Equations
Next, we add equations \(3\) and \(2\) to eliminate \(y\):
\[
30x - 6y + x + 6y = 78 + 15 \implies 31x = 93
\]
Solving for \(x\):
\[
x = \frac{93}{31} = 3
\]
Step 4: Substitute to Find \(y\)
Now that we have \(x\), we substitute \(x = 3\) back into one of the original equations to find \(y\). Using equation \(1\):
\[
5(3) - y = 13 \implies 15 - y = 13 \implies y = 2
\]
Final Answer
The solution to the system of equations is:
\[
\boxed{(3, 2)}
\]